Quote from: CRANKYpants on June 01, 2008, 08:38:49 AM
YOU NEED POWER (OR A STATOR CURRENT INCREASE) TO TAKE THE ROTOR FROM ONE RPM LEVEL TO A HIGHER ONE - TO ACCOUNT FOR THE INCREASED ENERGY STORED AS INERTIA.

That's right but you have not set up any way of measuring that temporary increase in power so you have no way of telling where it's supplied from. You are only measuring steady-state speed, current, power etc, not the transitionary values. The small power boost to supply the extra rotational kinetic energy only happens during the period of acceleration -  once the new speed is attained, it does not require any extra power to keep it going at the new speed (except for a very small increase in friction losses of course).

Quoteyou have not set up any way of measuring that temporary increase in power so you have no way of telling where it's supplied from. You are only measuring steady-state speed, current, power etc, not the transitionary values

HERE'S THE DEALIO,

A 0.01 A DECREASE IN STATOR CURRENT MEANS ACCELERATION AND COMPLIMENTARY TORQUE SUPPLIED TO THE MOTOR FROM THE COILS.

WE DON'T NEED TO KNOW WHERE IT IS COMING FROM - BUT WE CAN CONCLUDE WHERE IT IS NOT COMING FROM I.E. THE MOTOR BECAUSE...

THE MOTOR CURRENT WOULD HAVE TO INCREASE BY MORE THAN 0.01 AMPS TO CAUSE ACCELERATION AND THIS DOES NOT OCCUR.

WRONGIO - WE MEASURE STEADY STATE A AND TRANSITION AND STEADY STATE B

IN FACT THE TRANSITION PERIOD FROM OPEN HV COIL TO SHORTED HV COIL IS THE MOST IMPORTANT AND CRITICAL OBSERVATION TIME WHERE WE CONCLUDE:

A WORKING HV COIL = 0.01 A DECREASE IN STATOR CURRENT
A NON WORKING HV COIL = 0.01 A INCREASE IN STATOR CURRENT

THANEIOS

Quote from: CRANKYpants on June 01, 2008, 01:39:45 PM

A WORKING HV COIL = 0.01 A DECREASE IN STATOR CURRENT
A NON WORKING HV COIL = 0.01 A INCREASE IN STATOR CURRENT[/color][/b]

There are three states:

1. Before the acceleration
2. During the acceleration
3. After the acceleration

What happens to the current for each of these three states? You keep referring to only two.

Quote from: CRANKYpants on June 01, 2008, 09:11:49 AM
KEEP ADDING TURNS UNTIL YOU GET THIS:

Coil 3:   ?v    0.001 A  ?turns

THEN WE CAN DISCUSS - APPLES TO ORANGES.

Thane

My and Ron's HV coils are outputting far more that .001A

But you have not answered my question, but if you need me to restate here goes:

Coil1:       5v   1A      100 turns
Coil 2: 62.5v  .08A   1250 turns

I assure you each will have the same energy output and produce an equal magnetic field.
My question to you is if I or someone else can show parameters such as the above if you will accept that as evidence that slowing HC coils are not needed.

Or would I need to run the HV through a transformer so it is outputting close to the same amps and volts as the HC coil?

This isn't apples and oranges, it is all electrical power of the same energy and a magnetic field from each coil that assuming the coils are of the same volume and metal should be indistinguishable.

QuoteThere are three states:

1. Before the acceleration
2. During the acceleration
3. After the acceleration

What happens to the current for each of these three states? You keep referring to only two.

1) BEFORE ACCELERATION

WE TAKE THE MOTOR UP TO A STEADY STATE SPEED - THEN REDUCE THE VOLTAGE JUST ENOUGH TO ENSURE THAT THE SYSTEM IS DECELERATING EVER SO SLIGHTLY - SUCH THAT THE INDUCED VOLTAGE IN OUR COIL(S) IS DECREASING.

STATOR CURRENT IS STABLE.

THEN SHORT OUT THE HV COIL(S)

2) DURING ACCELERATION

ONCE THE HV COIL HAS BEEN ENGAGED - EITHER THE STATOR CURRENT INCREASES (DECELERATION) OR IT DECREASES (ACCELERATION)

IF THE HV COIL IS CAUSING ACCELERATION (NOT THE MOTOR) THE CURRENT WILL START TO DROP IMMEDIATELY - NOT EVEN A 0.01 AMP INCREASE IN STATOR CURRENT.

3) AFTER ACCELERATION

AFTER ACCELERATION HAS PEAKED THE HV COIL IS DISENGAGED AND IT IS IMPERATIVE THAT THE SYSTEM BEGINS TO DECELERATE WITH A CORRESPONDING INCREASE IN STATOR CURRENT - WHICH INDICATES THAT THE MOTOR IS NOT RESPONSIBLE FOR THE NEW ACCELERATED SYSTEM SPEED.

The Thane of Cawdor